6.2.2 Preparation of Equilibrium Mixtures and Equilibrium Constant You will prep

Question

6.2.2 Preparation of Equilibrium Mixtures and Equilibrium Constant You will prepare a new series of solutions that have varied concentrations of the Fe ions and the SCN ions You will use the results of this test to accurately evaluate the equilibrium concentrations of each species 1 2. Label (with tape) five white cups, 1-5, for the equilibrium sample mixtures. To prepare the sample solutions, use the same previously labeled 10.0 mL graduated pipet for transferring KSCN and a separate labeled 10.0 mL graduated pipet for transferring the HNO. 3. Rinse the buret with DI water and then about 10 mL of the 2.00 x 103 M Fe(NO)s. Then use the buret to transfer 10.00 mL of the 2.00 x 10 M Fe(NOsh into each beaker for the equilibrium sample solutions. Carefully pipet the appropriate amounts of KSCN and HNO, into each beaker, according to the Sample Solutions Table below. 4. 5. 6. Swirl each cup to mix the solutions. 7. In the Part B Data Table, measure and record the absorbance of the equilibrium solutions, 1-6 *Your instructor will provide instruction on this part of the exercise Table 3: Preparation of Equilibrium Solutions Solutions | Volume | 2.00 × 10' M Fe(NO)-KSON Solution | Volume of 0.5 M HNO3 8.00 mL 6.00 mL 4.00 mL Equilibrium Total Volume of Volume of 0,00 mL10.00 mL 2.00 mL 4,00 mL 6.00 mL 8.00 mL2.00 mL 10.00 mL 20.00 mL 10.00 220.00 mL10.00 mL 20.00 mL 20.00 mL 20.00 mL 20.00 mL 10.00 mL 10.00 mL 10.00 mL 10.00 mL 0.00 mL 6.2 9?

Explanation / Answer

From the given data,

molar absorptivity = 4295 M-1.cm-1

For expt. 1,

with no [SCN-] added

[FeSCN2+]eq = 0

For expt. 2,

[FeSCN2+]eq = (0.25 - 0.06)/4295 = 4.42 x 10^-5 M

initial [Fe3+] = 0.002 M x 10 ml/20 ml = 0.001 M

[Fe3+eq = 0.001 - 4.42 x 10^-5 = 9.56 x 10^-4 M

initial [SCN-] = 0.002 M x 2 ml/20 ml = 0.0002 M

[SCN-]eq = 0.0002 - 4.42 x 10^-5 = 1.56 x 10^-4 M

Keq = 4.42 x 10^-5/(9.56 x 10^-4 x 1.56 x 10^-4) = 296.4

For expt. 3,

[FeSCN2+]eq = (0.45 - 0.06)/4295 = 9.08 x 10^-5 M

initial [Fe3+] = 0.002 M x 10 ml/20 ml = 0.001 M

[Fe3+eq = 0.001 - 9.08 x 10^-5 = 9.09 x 10^-4 M

initial [SCN-] = 0.002 M x 4 ml/20 ml = 0.0004 M

[SCN-]eq = 0.0004 - 9.08 x 10^-5 = 3.09 x 10^-4 M

Keq = 9.08 x 10^-5/(9.09 x 10^-4 x 3.09 x 10^-4) = 323.0

For expt. 4,

[FeSCN2+]eq = (0.69 - 0.06)/4295 = 1.47 x 10^-4 M

initial [Fe3+] = 0.002 M x 10 ml/20 ml = 0.001 M

[Fe3+eq = 0.001 - 1.47 x 10^-5 = 8.53 x 10^-4 M

initial [SCN-] = 0.002 M x 6 ml/20 ml = 0.0006 M

[SCN-]eq = 0.0004 - 1.47 x 10^-4 = 4.53 x 10^-4 M

Keq = 1.47 x 10^-4/(8.53 x 10^-4 x 4.53 x 10^-4) = 380.4

So,

average Keq = 333.3

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